# Industrial Engineering Chapter 5 Homework The marginal probability distribution of X is normal with

Page Count
10 pages
Word Count
3206 words
Book Title
Applied Statistics and Probability for Engineers 7th Edition
Authors
Douglas C. Montgomery, George C. Runger
Applied Statistics and Probability for Engineers, 7th edition 2017
5-18
Section 5-5
5.5.1 a) percentage of slabs classified as high with probability p1 = 0.05
b) X is the number of voids independently classified as high X 0
5.5.2 a) Probability for the kth landing page = pk = 0.25
Applied Statistics and Probability for Engineers, 7th edition 2017
5-19
5.5.3 Because
= 0 and X and Y are normally distributed, X and Y are independent. Therefore,
5.5.4 a)
Applied Statistics and Probability for Engineers, 7th edition 2017
5-20
b)
c)
5.5.5 Because
= 0 and X and Y are normally distributed, X and Y are independent. Therefore,
X = 0.1 mm,
X = 0.00031 mm,
Y = 0.23 mm,
Y = 0.00017 mm
5.5.6 a)
= cov(X,Y)/
x
y = 0.6, cov(X,Y) = 0.6(2)(5) = 6
Applied Statistics and Probability for Engineers, 7th edition 2017
5.5.7

 

    −+


− − − −


==


   
2
2()
()
1
2 2 2
1
( , ) 2
Y
X
XY
y
x
XY
XY
f x y dxdy e dxdy
5.5.8 a) X1, X2 and X3 are binomial random variables when considered individually, that is, the marginal
probability distributions are binomial. Their joint distribution is multinomial.
= = =
1 1 2 2 3 3
( , , )
P X x X x X x
Section 5-6
5.6.1 a) E(3X + 2Y) = 3(2) + 2(6) = 18
5.6.3 a) X N(0.1, 0.00031) and Y N(0.23, 0.00017) Let T denote the total thickness.
5.6.4 a) X: time of wheel throwing. X ~ N(40,4)
5.6.5 X = time of ACL reconstruction surgery for high-volume hospitals.
5.6.6 a) Let
X
denote the average fill-volume of 100 cans
==
2
0.5 100 0.05
X
5.6.7 X ~ N(160, 900)
5.6.8 D = A B C
5-23
5.6.9
12
( ) (2 2 )
V Y V X X
=+
5.6.10 Let Y be the rate of return for the entire investment after one year,
5.6.11 Let Xi be the demand in month i.
a)
1 2 12
...
Z X X X
= + + +
Applied Statistics and Probability for Engineers, 7th edition 2017
5-24
Section 5-7
5.7.1
1
() 4
Y
fy=
at y = 3, 5, 7, 9
1
2
1
y
e
5.7.4 a) Now,
2
0
bv
av e dv
must equal one. Let u = bv, then
2
2
3
00
1uu
udu a
a e u e du
b
bb

−−

==



. From the
5.7.5 If y = ex, then x = ln y for 1 ≤ x ≤ 2 and e1 y e2. Thus,
11
( ) (ln )
YX
f y f y yy
==
for 1 ≤ ln y ≤ 2.
5.7.6
ln
W
Ye
WY
=
=
5.7.7
( )

= + + 0.5
22
1 2 1 2 1 2
(cos cos )rrrrr
Applied Statistics and Probability for Engineers, 7th edition 2017
5-25
Then, f(r) = f(
2)|J|, where
5.7.8 Here, X is lognormal with
= 5.2933, and
2 = 0.00995
Y = X4
Section 5-8
5.8.1 a)
−−
+
= = =
= = = = =
 
11
( 1)
1 0 0
1 1 (1 )
( ) ( ) (1 )
t t tm
m m m
tX tx t x tx
Xt
x x x
e e e
M t E e e e e
m m m me
Applied Statistics and Probability for Engineers, 7th edition 2017
5-26
5.8.3 a)

==
= = − =

1
11
( ) ( ) (1 ) ( (1 ))
1
tX tx x t x
X
xx
p
M t E e e p p e p
p
b)
( ) ( )
( )

=
=
− −
= = = = −−
12
0
0
1 (1 ) (1 )
()
( ) '
1 (1 )
t t t t
X
t
t
t
pe p e pe p e
dM t
EX dt pe
5.8.5 a)

−−
==
= = =

2 ( 2)
00
( ) ( ) (4 ) 4
tX tx x t x
X
xx
M t E e e xe dx xe dx
Applied Statistics and Probability for Engineers, 7th edition 2017
5-27
5.8.6
a)
 
       
=
=
 
= = = = − =
 
− −
 
1 1 1
( ) ( ) ()
x
tx t t t t
tX tx
X
x
e e e e e
M t E e e dx t t t t
( ) ( )
 
 

+ −
=
2 2 2
23
0
2 2 2
' lim ()
t t t t t t
t
t e e t e e e e
t
5.8.7 a)



−−
= = =

()
( ) ( )
tX tx x t x
X
M t E e e e dx e dx
Applied Statistics and Probability for Engineers, 7th edition 2017
5-28
5.8.8 a)
( )



− −
= = =


11 ( )
00
( ) ( ) ( ) ( )
r
r
tX tx x r t x
X
M t E e e x e dx x e dx
rr
−−
1 ( )
0
r t x
x e dx
is finite only if t < λ. Besides, we need to use integration by substitution by
a)
( ) ( )
 

− −
==
==

= = = = = = =


1
100
00
()
( ) ' 1
rrr
rr
X
tt
tt
dM t t r
E X t r t
dt
5.8.9 a)
 
 

= = = 
− −

12
( ) ( ). ( )... ( ) ...
r
r
Y X X X
M t M t M t M t t t t t
5.8.10 a)


 
= = +  +
 
 
12
2 2 2 2
12
12
( ) ( ) ( ) exp exp
22
Y X X
tt
M t M t M t t t
Applied Statistics and Probability for Engineers, 7th edition 2017
5-29
Supplemental Exercises
5.S11 The sum of
=
 ( , ) 1
xy
f x y
,
         
+ + + + =
         
         
1 1 1 1 1 1
4 8 8 4 4
and fXY(x,y) ≥ 0
5.S12 a)
= = = = =
2 4 14
20!
( 2, 4, 14) 0.10 0.20 0.70 0.0631
P X Y Z
Applied Statistics and Probability for Engineers, 7th edition 2017
5-30
5.S13
= = =
 
3 2 3 23
23
22
00
0 0 0
2 18
23
yx
cx ydydx cx dx c c
. Therefore, c = 1/18.
5.S14 The region x2 + y2 ≤ 1 and 0 < z < 4 is a cylinder of radius 1 (and base area π) and height 4.
Therefore, the volume of the cylinder is 4π and
=1
( , , ) 4
XY Z
f x y z
for x2 + y2 ≤ 1 and 0 < z < 4.
Applied Statistics and Probability for Engineers, 7th edition 2017
5-31
5.S15 Let X, Y, and Z denote the number of problems that result in functional, minor, and no defects,
respectively.
c) E(Z) = 10(0.3) = 3
5.S16 a) Let
X
denote the mean weight of the 25 bricks in the sample. Then,
=( ) 3E X
and
5.S17 Let
X
denote the average time to locate 10 parts. Then,
=4( 5)E X
and
=30
10
X
5.S18 a) Let X denote the weight of a piece of candy and X N(0.1, 0.01). Each package has
Applied Statistics and Probability for Engineers, 7th edition 2017
5-32
5.S19
5.S20 a) Let Y denote the weight of an assembly. Then, E(Y) = 4 + 5.5 + 10 + 8 = 27.5 and V(Y) = 0.42 +
5.S21 Let T denote the total thickness. Then, T = X1 + X2 and
5.S22 Let X and Y denote the percentage returns for security one and two, respectively.
If half of the total dollars is invested in each, then 1/2X + 1/2Y is the percentage return.
5.S23 a) Let X, Y, and Z denote the risk of new competitors as no risk, moderate risk,
Applied Statistics and Probability for Engineers, 7th edition 2017
5-33
5.S24
Y X X Y
==
Then
5.S25
2
I
P I PR
R
==
5.S26 Because the covariance is zero, one may consider a cross section of the beam, say along the
x axis. The distribution along the x axis is

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